Ural 1416 Confidential（最小生成树+次小生成树）

来源：互联网发布：nginx ip映射域名编辑：程序博客网时间：2024/06/11 08:21

题意：给n个点m条边，求它的最小生成树和次小生成树的权值分别是多少。

思路：先求最小生成树，并查集即可，求出来之后求次小生成树，次小生成树一定是在最小生成树的基础上添加一条边在之后形成的环中删除除了加入的边中的最大的一条边，然后枚举所有未添加的边。求两点之间路径上的最大权的边，可以预处理出LCA，然后查询log n，就可以了。

#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<set>#include<algorithm>const int maxn = 1e3 + 10;const int INF = 1e9;using namespace std;struct P {    int u, v, cost;    P() {}    P(int u, int v, int c) : u(u), v(v), cost(c) {}    bool operator < (P p) const { return cost < p.cost; }    void input() { scanf("%d %d %d", &u, &v, &cost); }} tr[maxn * maxn];int n, m, pre[maxn], c[maxn];int use[maxn * maxn];int vis[maxn], p[maxn], deep[maxn];int anc[maxn][30], cost[maxn][30];vector<P> G[maxn];void init() {    for(int i = 0; i < maxn; i++) {        G[i].clear();        pre[i] = i;    }    memset(use, 0, sizeof use);    memset(vis, 0, sizeof vis);}int findset(int x) { return pre[x] = (x == pre[x] ? x : findset(pre[x])); }void dfs(int x, int k) {    vis[x] = 1; deep[x] = k;    for(int i = 0; i < G[x].size(); i++) {        int v = G[x][i].v;        if(vis[v]) continue;        p[v] = x;  dfs(v, k + 1);        c[v] = G[x][i].cost;    }}void preprocess() {    for(int i = 1; i <= n; i++) {        anc[i][0] = p[i]; cost[i][0] = c[i];        for(int j = 1; (1 << j) < n; j++) anc[i][j] = -1;    }    for(int j = 1; (1 << j) < n; j++) {        for(int i = 1; i <= n; i++) {            if(anc[i][j - 1] == -1) continue;            int a = anc[i][j - 1];            anc[i][j] = anc[a][j - 1];            cost[i][j] = max(cost[i][j - 1], cost[a][j - 1]);        }    }}int query(int p, int q) {    int lg;    if(deep[p] < deep[q]) swap(p, q);    for(lg = 1; (1 << lg) <= deep[p]; lg++); lg--;    int ans = 0;    for(int i = lg; i >= 0; i--) {        if(deep[p] - (1 << i) >= deep[q]) {            ans = max(ans, cost[p][i]); p = anc[p][i];        }    }    if(p == q) return ans;    for(int i = lg; i >= 0; i--) {        if(anc[p][i] != -1 && anc[p][i] != anc[q][i]) {            ans = max(ans, cost[p][i]); p = anc[p][i];            ans = max(ans, cost[q][i]); q = anc[q][i];        }    }    ans = max(ans, max(c[p], c[q]));    return ans;}int main() {    while(scanf("%d %d", &n, &m) != EOF) {        init();        for(int i = 0; i < m; i++) tr[i].input();        sort(tr, tr + m);        int ans1 = 0, ans2 = INF, edge = 0;        for(int i = 0; i < m; i++) {            int u = tr[i].u, v = tr[i].v;            int nx = findset(u), ny = findset(v);            if(nx == ny) continue;            pre[nx] = ny;  ans1 += tr[i].cost;            edge++;        use[i] = 1;            G[u].push_back(P(u, v, tr[i].cost));            G[v].push_back(P(v, u, tr[i].cost));        }        if(edge < n - 1) ans1 = -1;        if(ans1 != -1) {            p[1] = -1; c[1] = 0;            dfs(1, 1);            preprocess();            for(int i = 0; i < m; i++) {                if(use[i]) continue;                int u = tr[i].u, v = tr[i].v;                int add = query(u, v);                ans2 = min(ans2, ans1 + tr[i].cost - add);            }        }        if(ans2 == INF) ans2 = -1;        printf("Cost: %d\nCost: %d\n", ans1, ans2);    }    return 0;}

0 0